Given an array of n elements.You can double or triple the elements in the array any number of times. After all the operations check whether it is possible to make all elements in the array equal.
Examples :
Input : A[] = {75, 150, 75, 50}
Output : Yes
Explanation : Here, 75 should be doubled twice and
150 should be doubled once and 50 should be doubled
once and tripled once.Then, all the elements will
be equal to 300.
Input : A[] = {100, 151, 200}
Output : No
Explanation : No matter what we do all elements in
the array could not be equal.
The idea is to repeatedly divide every element by 2 and 3 until the element is divisible. After this step, if all elements become same, then answer is yes.
How does this work? We know that every integer can be expressed as product of prime numbers 2a.3b.5c.7d…... So, in our problem we can increase a and b by doubling(*2) or tripling(*3). We can make a and b of all elements in the array equal by multiplying with 2 or 3. But the numbers also have other prime numbers in their product representation, we cannot change the powers of them. So to make all numbers equal they should have powers on other prime numbers equal from the beginning. We can check it by dividing all the numbers by two or three as many times as possible. Then all of them should be equal.
Implementation:
C++
// C++ program to check if all numbers can// be made equal by repeated division of 2// and 3#include <bits/stdc++.h>using namespace std;bool canMakeEqual(int a[], int n){ for (int i = 0; i < n; i++) { // continuously divide every number by 2 while (a[i] % 2 == 0) a[i] = a[i] / 2; // continuously divide every number by 3 while (a[i] % 3 == 0) a[i] = a[i] / 3; } // Check if all numbers same for (int i = 1; i < n; i++) if (a[i] != a[0]) return false; return true;}// Driver Codeint main(){ int A[] = { 75, 150, 75, 50 }; int n = sizeof(A) / sizeof(A[0]); if (canMakeEqual(A, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to check if all numbers can// be made equal by repeated division of 2// and 3import java.util.*;class GFG {static Boolean canMakeEqual(int a[], int n){ for (int i = 0; i < n; i++) { // Continuously divide every number by 2 while (a[i] % 2 == 0) a[i] = a[i] / 2; // Continuously divide every number by 3 while (a[i] % 3 == 0) a[i] = a[i] / 3; } // Check if all numbers same for (int i = 1; i < n; i++) if (a[i] != a[0]) return false; return true;}// Driver Codepublic static void main(String[] args){int A[] = { 75, 150, 75, 50 }; int n = A.length; if (canMakeEqual(A, n)) System.out.print("Yes"); else System.out.print("No");} }// This code is contributed by 'Gitanjali'. |
Python3
# Python3 code to check if all numbers can# be made equal by repeated division of 2# and 3def canMakeEqual( a , n ): for i in range(n): # continuously divide every number by 2 while a[i] % 2 == 0: a[i] = int(a[i] / 2) # continuously divide every number by 3 while a[i] % 3 == 0: a[i] = int(a[i] / 3) # Check if all numbers same for i in range(1,n): if a[i] != a[0]: return False return True # Driver CodeA = [ 75, 150, 75, 50 ]n = len(A)print("Yes" if canMakeEqual(A, n) else "No")# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to check if all numbers can// be made equal by repeated division of 2// and 3using System;class GFG { static Boolean canMakeEqual(int []a, int n) { for (int i = 0; i < n; i++) { // Continuously divide every number by 2 while (a[i] % 2 == 0) a[i] = a[i] / 2; // Continuously divide every number by 3 while (a[i] % 3 == 0) a[i] = a[i] / 3; } // Check if all numbers same for (int i = 1; i < n; i++) if (a[i] != a[0]) return false; return true; } // Driver Code public static void Main() { int []A = { 75, 150, 75, 50 }; int n = A.Length; if (canMakeEqual(A, n)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by 'vt_m'. |
PHP
<?php// PHP program to check if // all numbers can be made // equal by repeated division // of 2 and 3function canMakeEqual($a, $n){ for ($i = 0; $i < $n; $i++) { // continuously divide // every number by 2 while ($a[$i] % 2 == 0) $a[$i] = $a[$i] / 2; // continuously divide // every number by 3 while ($a[$i] % 3 == 0) $a[$i] = $a[$i] / 3; } // Check if all numbers same for ($i = 1; $i < $n; $i++) if ($a[$i] != $a[0]) return false; return true;}// Driver Code$A = array(75, 150, 75, 50);$n = sizeof($A);if (canMakeEqual($A, $n)) echo "Yes";else echo "No";// This code is contributed by aj_36?> |
Javascript
<script>// Javascript program to check if all numbers can// be made equal by repeated division of 2// and 3function canMakeEqual(a, n){ for(let i = 0; i < n; i++) { // Continuously divide every number by 2 while (a[i] % 2 == 0) a[i] = a[i] / 2; // Continuously divide every number by 3 while (a[i] % 3 == 0) a[i] = a[i] / 3; } // Check if all numbers same for(let i = 1; i < n; i++) if (a[i] != a[0]) return false; return true;}// Driver codelet A = [ 75, 150, 75, 50 ];let n = A.length;if (canMakeEqual(A, n)) document.write("Yes");else document.write("No"); // This code is contributed by suresh07</script> |
Yes
Time complexity : O(n * log(max(A))), where n is the size of the input array A and max(A) is the maximum value in the array.
Space complexity : O(1),
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